Question: Factor the quadratic expression completely. $3x^2-20x-7=$
Answer: Since the terms in the expression do not share a common monomial factor and the coefficient on the leading $x^2$ term is not $1$, let's factor by grouping. The expression ${3}x^2{-20}x{-7}$ is in the form ${A}x^2+{B}x+{C}$. First, we need to find two integers ${a}$ and ${b}$ such that: $\begin{cases} &{a}+{b}={B}={-20} \\\\ &{ab}={A}{C}= ({3})({-7})=-21 \end{cases}$ We find that ${a}={1}$ and ${b}={-21}$ satisfy these conditions, since ${1}+({-21})={-20}$ and $({1})({-21})=-21$. Next, we can use these values to rewrite the $x$ -term and factor by grouping. $\begin{aligned} 3x^2-20x-7&=3x^2+{1}x{-21}x-7 \\\\ &=x(3x+1)-7(3x+1) \\\\ &=(3x+1)(x-7) \end{aligned}$ In conclusion, $3x^2-20x-7=(3x+1)(x-7)$